Deciding
Linear-Trigonometric
Problems
FUJITSU LABORATORIES LTD
穴井
宏和
(Hirokazu
ANAI)
$*$universit\"at
$\mathrm{P}\mathrm{a}\mathrm{S}\mathrm{S}\mathrm{a}\mathrm{u}$Volker WEISPFENNING\dagger
Abstract
In thispaper, wepresentadecision procedure forcertainlinear-trigonometric problems for the reals andintegersformalizedinasuitable first-order language. Theinputs arerestrictedto formulas,
whereall butoneofthequantified variablesoccurlinearly andat mostoneoccursbothlinearlyand
in a specific trigonometric function. Moreoverwe allow in addition the integer-part operation in formulas. Besides ordinary quantifiers, we allow also countingquantifiers. Furthermore we also determine the qualitativestructureof the connectedcomponentsofthesatisfactionsetofthemixed linear-trigonometric variable. We alsoconsiderthedecision ofthese problems in subfields ofthe
real algebraic numbers.
1
Introduction
Decision and quantifier elimination methods for the real numbers as ordered field have a
large variety of applications in mathematics, computer science, and industrial engineering (see
$e.g$. $[3,25])$. In particular, many engineering and industrial applications
can
be reduced to theQEproblems containingtrigonometricfunctions(see [9, 6]).
Unfortunately, the general first-order theory of the real numbers with sine or cosine and
the constant $\pi$ is undecidable: this is because the zeros of sine function are exactly the integer
multiples of$\pi$. As a consequence one can code Hilbert’s tenth problem in existential formulas
$\mathrm{i}\mathrm{n}\mathrm{v}\mathrm{o}]\mathrm{v}\mathrm{i}\mathrm{n}\mathrm{g}+,$
$-,$$\cdot$, the sinefunction andtheconstants $0,1,$$\pi$ (compare [13]). In fact as remarked
already in [19] even the first-order theory ofthe real numbers with sine (without the constant
$\pi)$ is undecidable, since one can code the first-order theory ofrational numbers in this theory
by representing rational numbers as quotients of arbitrary
zeros
of the sine function. In thespecial case, where the quantified variables of the given formula split into purely polynomial
ones
andpurelytrigonometricones
the decisionproblemand the quantifiereliminationproblemcan
be reducedto the theory of real numbers as ordered field. It suffices to replace for apurelytrigonometricvariable$x$theterms $\sin(x)$ and$\cos(x)$ bynewvariables $u,$$v$, respectively, with the
additional relation $u^{2}+v^{2}=1$ (compare [7,4,5]). Theintroduction oftwo
new
variables inplaceof
one
trigonometric variables creates, however, serious problems in practical computationalcomplexity. A
more
efficient alternativewas
proposed,
by Pau et al.; it adapts Collins’ CADmethod [2] to the trigonometric case [12]. And trigonometric problems of special types have
been studied byD.Richardson by avariety of approaches [14, 15, 16, 17].
Here we consider the decision problem for a fragment of the full first-order theory of the
reals
as
ordered field withone
specific trigonometric function. In contrast to the special case,wherequantified variables
are
separated into polynomial and trigonometric variables,we
allowone
mixedquantified variable$x$ occurringboth in the trigonometric functionand in linearpoly-nomials. All other quantified variables $x_{i}$ may
occur
only linearly. Parametersmay occur
inarbitrary degree. We
assume moreover
that the mixed variable $x$ is quantified last (i.e.outer-most) inthegiven formula. Closed formulas of this restrictedtype
are
called normal. Wepresentan decision procedure for normal formulas. The procedurehas certain parallels in its approach
tothe decision procedure for linear-exponential problems announced in [26]. In fact bya simple
coding trickwe could also allow awhole block of quantifierswith respect to mixed variables in
place of
a
single mixedvariable.Both extend also to the mixed real-integer case, $i.e$, to formulas, where terms may involve
the integer-part operation and henceimplicitly quantifiers
over
the integers. Our resultsmay beviewed
as
afirst step towards decidable classes ofproblemswith mixed algebraic-trigonometricreal andintegervariables thataremotivatedbythe study ofcontroltheoryofhybridsystems [8].
The proofs involve real and mixedreal-integerlinear quantifier elimination [21, 23, 10, 27],
elementary analysis, the idea of checking the real satisfiability of formulas by finitely many
symbolic test points, and Lindemann’s theorem from transcendental number theory in orderto
decide relations between constant terms. In contrast to the purely linear
case
considered in$[10, 27]$ thetestpointsconsideredherehaveno straightforward mathematicalrepresentation. We
inventasymbolic notationforthem,and show,howto sortthesepoints intheirnatural order,and
how to evaluate the sign of other terms at these symbolically represented points in an indirect
manner.
We also solve the decision problem for normal formulas, where quantifiers range
over
anarbitrary subfield$F$of the field of real algebraic numbers. In contrast topurely linearproblems,
mixed linear-trigonometric problems mayhave different truth values overthe reals and over$F$.
Besides ordinary quantifiers, we mayalso allowcounting quantifiers.
In contrast to the general decision procedure for the first order theory of the reals as
ex-ponential field that relies on Schanuel’s conjecture [11], we require no open conjecture from
transcendence theory. Instead we use only Lindemann’s theorem and the transcendency of$\pi$ in
orderto decide relations betweenconstants.
Theclassicaltheorem of Lindemann(see [18],page 70)asserts that the values of thecomplex
exponential functionfor
non-zero
complex algebraic argumentsare
transcendental. Asa
conse-quence acorresponding assertion holds forreal trigonometric functions
as
well. This is seenas
follows: Call areal orcomplex function $f(z)$ is stronglytranscendental(withexceptional point
$p)$ if for all real
or
complex numbers $(z\neq p)$ it is not the case thatboth $z$ and $f(z)$ arealge-braic. Then by Lindemann’s theorem $e^{z}$ is strongly transcendental with exceptionalpoint $0$. As
an immediate consequence we see that the real logarithm $\ln(x)$ is strongly transcendental with
exceptionalpoint 1.
Next weclaim that the real sinus and cosinus functions are stronglytranscendental with
ex-ceptional point $0$. Indeed $e^{ix}=\cos(x)+i\sin(x)$ and $\cos^{2}(x)+\sin^{2}(X)=1$. So if $\cos(x)$
is algebraic, then by the second equation $\sin(x)$ is algebraic; since $i$ is algebraic, this implies
that $i\sin(x)$, and hence $e^{ix}$ is algebraic. By Lindemann’s theorem this implies that $ix$, and
$\arcsin(X)$
are
also strongly transcendental with exceptionalpoint 1 and $0$, respectively. In orderto
see
that the real tangent function is strongly transcendental with exceptionalpoint $0$,we
usethewell-known formula forthe arcustangent: $\arctan(x)=-\frac{i}{2}\ln(\frac{1+ix}{1-ix})$. It shows thatfor
alge-braic $x\neq 0\arctan(X)$ is transcendental. As
a
consequence $\arctan(x)$ and$tan(X)$are
stronglytranscendental with exceptional point $0$. In a similar way
one
sees that the real arcuscotangentfunction and hencethe real cotangent function are strongly transcendental without exceptional
point.
2
The
formal framework
Let $F$ be an arbitrary subfield $F$ of the field of real algebraic numbers. We consider the
following fragment $L$ ofa first-order language for the real numbers. Let $x$ be a distinguished
variable. Lineartermsareexpressions of the form $s:=cx+ \sum_{i=1}^{n}C_{i}X_{i}+c_{0}$with$c;c_{i}\in F$. The
trigonometric term is
an
expressions of the form $t:=trig(ax+b)$, where trig denotes a realtrigonometric function sine, cosine or tangent, and $a(\neq 0),$ $b$ are fixed elements of$F$. Atomic
formulas are equations or inequalities of the form $s=0,$ $s=t,$$s<t,$$t<s$, where $s$ is a
lineartermand$t$isantrigonometricterm. Arbitraryformulas
are
obtained fromatomic formulasas usual in first order logic, $i.e$. by iterated application ofthe propositional operations $\mathrm{A},$ $,$$\neg$
andquantifiers $\exists x,$$\exists x_{i},$$\forall X,$ $\forall x_{i}$ with respectto the variable$x$
or
tosome
$x_{i}$. Every formula can
be rewritten equivalently in prenex normal form. If all
occurrences
of $x$ in a formula $\varphi$ arequantified, then $\varphi$ is a closed
formula.
In the natural interpretation of formulas over the realnumbers,every closed formula is eithertrue orfalse. We call aclosed formula $\varphi$ normal ifit is
prenexand either doesnotcontainthevariable$x$orthe quantifier$\exists x$or$\forall x$binding$x$isoutermost
in theprefix of$\varphi$. Ourgoal is a decision procedure for normal formulas in $L$ i.e. an algorithm
that takes normal formulas in $L$ as inputs and Output the truth value ofthe input formula over
the reals. This goal is analogoustothat in [22] foralmost linearinteger problems.
In fact we could allow in normal formulas a whole block of quantifiers referring to mixed
variables, say$y_{1},$ $\ldots$
,
$y_{m}$ in placeofasingle mixedvariable$x$. Thentrigonometric terms wouldbeoftheform trig$(a_{0}+a_{1}y_{1}+.. . a_{m}y_{m})$ forfixedreal constants$a_{0},$
. .
${ }$ ,$a_{m}$. Suchaformula$\exists y_{1}\ldots\exists ymQ1x_{1}\ldots Qnxn(\psi)$ witharbitrary quantifiers $Q_{i}$ canbe equivalently replacedby the
followingnormalformula
$\exists x\exists y_{1}\ldots\exists y_{m}Q_{1}X1\cdots Qn^{X}n$($x=a_{0}+a_{1}y_{1}+\ldots a_{m}y_{m}$ A $\psi’$),
where$\psi’$ results from$\psi$by replacingeach
occurence
oftrig$(a0+a_{1}y_{1}+\ldots a_{m}y_{m})$ by trig$(x)$.In fact we will consider also formulas in an extended sense, where we admit besides the
“ordinary” quantifiers $\exists,$$\forall$alsothe“counting quantifiers” $\exists\leq n,$$\exists^{n},$$\exists\geq n,$ $\exists\leq\infty,$$\exists^{\infty}$ interpreted
as
“thereexistatmost $n$, exactly$n$,atleast$n$, only finitely
many,
infinitely many”, respectively for$n\in \mathrm{N}$. Formulas,closedformulas, normalformulasinthis extended
sense
willbe referredtoasformulas,closed formulas, normalformulas with countingquantifiers. Notice thatthe first three
types ofcountingquantifiers are definable using ordinary quantifiers, while the lasttwo are not
definable in this way.
Let $\varphi:=Qx(\psi(x))$ be a normal formula ($Q$ a quantifier $\exists$ or $\forall$). Let
$\psi_{1}$ result from $\psi$
by replacing the trigonometric term trig$(ax+b)$ by a
new
variable $z$. Then $\psi_{1}$ is a linearformula in the
sense
of $[21, 10]$. So by linear real quantifier elimination, $\psi_{1}$ is equivalent to a$\psi_{2}$ bybacksubstitution of the trigonometric term $t=trig(ax+b)$ for $z$. Then $\varphi’:=Qx(\psi_{3})$
is
a
normal formula with onlyone
quantifierthat is equivalent to $\varphi$. Notice that rational linearcombinations oftrigonometric terms can be rewritten as arational multiple of a trigonometric
term. Sobydividing byrational constants theatomic formulas of$\varphi’$ canassumedtobe normal.
For normal formulas with counting quantifiers one can argue similarly, using linear
elimi-nation by test points. So for example
a
counting quantifier $\exists^{\infty}x_{i}(\sigma(X_{i}))$can
be eliminated byrestrictingthe disjunction
over
testtermsused forthe elimination of the quantifier $\exists x_{i}(\sigma(X_{i}))$ totest terms of the form $s\pm\epsilon$, where $s$is alinearterm.
So the problem of deciding the truth value of normal formulas is reduced to normal input
formulas of the form $Qx(\psi)$, where $\psi$ is quantifier-free. Moreover by passing to the negation
-ifnecessary-we may
assume
that $Q$ is an existential quantifier. So it suffices to decide suchunivariatenormal formulaswith a single existential quantifier $\exists x$
over
the reals. In factwe willdo
more:
Incase thetruth value of$\exists x(\psi)$ is ‘true’,we
will provide a sample point $r\in \mathbb{R}$for $x$.In fact if$M$ isthe set of reals satisfying $\psi(x)$, then we will determine the numberofconnected
components of$M$ and their types as anon-empty intervals $[b, c],$$(b, c),$ $[b, c)$ or $(b, c]$.
As a consequence we canalso inan obviousway decidea univariatenormal formula$Qx(\psi)$
with quantifier-free$\psi$, where$Q$ isacounting quantifier.
3
Geometric
properties of
trigonometric
functions
and
straight
lines
In ordertostudy the solutions ofequationsandinequalitiesbetweenstraightlines and
trigono-metric functions,we nowconsiderfunctionsof the type$f(x)=trig(x)-(cx+d)$,andfocus
on
the caseoftrig$(x)=\cos(x)$ or trig$(x)=\sin(x)$. In thefollowing, we only deal with the
case
of$\cos(x)$ because thecase of$\sin(x)$ is very analogous tothat of$\cos(x)$. Let the basic intervals
for be $\mathcal{I}_{n}:=(-\frac{\pi}{2}+n\pi, -\frac{\pi}{2}+(n+1)\pi)$ for $n\in$ N. For any straight line $y=cx+d$ the
equation $\cos(x)=cx+d$has at most two solutionsinthe interval$\mathcal{I}_{n}$. We denote thesesolutions
(iftheyexist) by $\xi_{n,1}$ and$\xi_{n,2}$.
Definition 1 Consider the equation $\cos(x)=cx+d$ with real numbers $c,$$d$ andassume that
$c>0$. Then let$x_{+1}= \frac{1-d}{c},$ $x_{0}=- \frac{d}{c}$ and $x_{-1}= \frac{-1-d}{c}$. Then we call the region $[x_{-1}, x_{+}1]$
from
$x_{-1}$ to $x_{+1}$ the domainof
the line $cx+d$, in particular, we call the region $[x_{01}, x_{+}]$from
$x_{0}$ to $x_{+1}$ the positive domain and the region $[x_{-1}, x_{0}]$
from
$x_{-1}$ to $x_{0}$ the negative domainof
the line. For$c<0$the
definitions
are analogous.Lemma2 Considerthe equation $\cos(x)=cx+d$ with real numbers $c,$$d$. Then the following
hold:
1.
If
$c=0and|d|<1$, then theequation has infinitely many solutions.$If1>d\geq 0$thenthe solutionsare$\xi_{n,1},\xi_{n,2},$ $where- \frac{\pi}{2}+n\pi\leq\xi_{n,1}<\xi_{n,2}\leq-\frac{\pi}{2}+(n+1)\pi$
for
alleven integer$n$. Moreover$\cos(x)\geq cx+d$for
$\xi_{n,1}\leq x\leq\xi_{n,2}$, otherwise $\cos(x)<$$cx+d$ in$\mathcal{I}_{n}$.
If $0>d>-1$
then the solutionsare $\xi_{n,1},\xi_{n,2}$ where $- \frac{\pi}{2}+n\pi\leq\xi_{n,1}<\xi_{n,2}\leq-\frac{\pi}{2}+$$(n+1)\pi$
for
all odd integer$n$. Moreover $\cos(x)\leq cx+d$for
$\xi_{n,1}\leq x\leq\xi_{n,2}$, otherwise2.
If
$c=0and|d|=1$, then the equation has infinitely many solutions $n\pi$ where $n$is evenfor
$d=1$ and odd
for
$d=-1$. Moreover $\cos(x)\geq cx+d$for
$d=-1,$ $\cos(x)\leq cx+d$for
$d=1$.
3.
If
$c=0$ and $|d|>1$, then the equation has no solution. Moreover $\cos(x)>cx+d$for
$d<-1,$ $\cos(X)<cx+d$
for
$d<1$.4. $If|c|\geq 1$ then the equation has exactlyone solution $\xi$. When $c\geq 1$ then $\cos(x)\geq cx+d$
for
$x\leq\xi,$ $\cos(X)<cx+d$for
$x>\xi$. And when $c\leq-1$ then $\cos(x)\geq cx+d$for
$x\geq\xi$, $\cos(x)<cx+d$for
$x<\xi$.5.
If
$0<c<1$
then thefollowing cases occurin the positive domain where $m,$$n$are integers$s.t$. $m<n.\cdot$
(i) $If- \frac{\pi}{2}+2m\pi<x_{0}<-\frac{\pi}{2}+(2m+1)\pi and-\frac{\pi}{2}+2n\pi<x_{+1}<2n\pi$ then the equation
has the solutions $\xi_{2m,1}<\xi_{2m+2,1}<\xi_{2m+,2}<\cdots<\xi_{2n-2,1}<\xi_{2n-2},2$.
Moreover $\cos(x)\geq cx+d$
for
$x_{0}\leq x\leq\xi_{2m,1},$ $\cos(x)<cx+d$for
$\xi_{2m,1}<x<$$\xi_{2m+2,1},$ $\cos(x)\geq cx+d$
for
$\xi_{2m+2,1}\leq x\leq\xi_{2m+2,2},$ $\cdots,$ $\cos(x)\geq cx+d$for
$\xi_{2n-2},1\leq x\leq\xi_{2n-2,2},$ $\cos(x)<cx+d$
for
$\xi_{2n-2,2}\leq x_{+1}$.(ii) $If- \frac{\pi}{2}+2m\pi<x_{0}<-\frac{\pi}{2}+(2m+1)\pi$and $2n \pi<x_{+1}<-\frac{\pi}{2}+(2n+1)\pi$ then the
equation has the solutions $\xi_{2m,1}<\xi_{2m+2,1}<\xi_{2m+2,2}<\cdots<\xi_{2n-2},1<\xi_{2n-2},2<$
$\xi_{2n,1}<\xi_{2n,2}$.
Moreover $\cos(x)\geq cx+d$
for
$x_{0}\leq x\leq\xi_{2m,1},$ $\cos(x)<cx+d$for
$\xi_{2m,1}<x<$$\xi 2m+2,1,$ $\cos(x)\geq cx+d$
for
$\xi_{2m+2,1}\leq x\leq\xi_{2m+2,2},$ $\cdots,$ $\cos(x)\geq cx+d$for
$\xi_{2n-2,1}\leq x\leq\xi_{2n-2,2},$ $\cos(x)<cx+d$
for
$\xi_{2n-2,2}<x<\xi_{2n,1},$ $\cos(X)\geq cx+d$for
$\xi_{2n,1}\leq x\leq\xi_{2n,2},$ $\cos(X)<cx+d$
for
$\xi_{2n,2}<x\leq x_{+1}$.(iii) $If- \frac{\pi}{2}+(2m+1)\pi<x_{0}<-\frac{\pi}{2}+(2m+2)\pi and-\frac{\pi}{2}+2n\pi<x_{+1}<2n\pi$ then the
equationhas the solutions$\xi_{2m+2,1}<\xi_{2m+2,2}<\cdots<\xi_{2n-2,1}<\xi_{2n-2,2}$.
Moreover $\cos(x)<cx+d$
for
$x_{0}<x<\xi_{2m+2,1},$ $\cos(x)\geq cx+d$for
$\xi_{2m+2,1}\leq$$x\leq\xi_{2m+2,2},$ $\cdots,$ $\cos(x)\geq cx+d$
for
$\xi_{2n-2,1}\leq x\leq\xi_{2n-2},2,$ $\cos(x)<cx+d$for
$\xi_{2n-2,2}\leq x_{+1}$.
(vi) $If- \frac{\pi}{2}+(2m+2)\pi<x_{0}<-\frac{\pi}{2}+(2m+2)\pi$ and$2n \pi<x_{+1}<-\frac{\pi}{2}+(2n+1)\pi$then
theequationhas the solutions$\xi_{2m+2,1}<\xi_{2m+2,2}<,$ $..<\xi_{2n-2},1<\xi_{2n-2}$}$2<\xi_{2n,1}<$
$\xi_{2n,2}$,
Moreover $\cos(x)<cx+d$
for
$x_{0}<x<\xi_{2m+2,1},$ $\cos(x)\geq cx+d$for
$\xi_{2m+2,1}\leq$$x\leq\xi_{2m+2,2},$ $\cdots,$ $\cos(x)\geq cx+d$
for
$\xi_{2n-2},1\leq x\leq\xi_{2n-2},2,$ $\cos(x)<cx+d$for
$\xi_{2n-2},2<x<\xi_{2n,1},$ $\cos(x)\geq cx+d$
for
$\xi_{2n,1}\leq x\leq\xi_{2n,2},$ $\cos(x)<cx+d$for
$\xi_{2n,2}<x\leq x_{+1}$.
Forthe negative domain wehavean analogouscasedistinction.
6.
$If- l<c<0$
then wehave analogous subcasesas incase5.4
Decision Procedures for
univariate
formulas
In this section we decide thevalidity ofunivariateclosed normal formulas inthelanguage $L$
ofthe form $\exists x(\psi)$, where $\psi$ is a quantifier-free. In view ofthe results in the section 2 this will
solvethe general decision problems for normal formulas.
Let the input formula $\varphi$ be of the form $\exists x(\psi(X))$, where $\psi$ is quantifier-free. The only
trigonometric termoccurring in$\psi$is trig$(ax+b)$where$a,$ $b$
are
elementsof$F$and$a\neq 0$. By thelinear substitution$x\vdasharrow a^{-1}x-a^{-}1b,$ $\varphi$is equivalentto$\exists x(\psi;)$, where$\psi’:=\psi(a^{-1}x-a^{-1}b))$.
Notice that the only trigonometric term occuring in $\psi’$ is trig$(x)$. So $\psi’$ is a propositional
combination ofatomicformulas
$\alpha_{i}:=c_{i}x+d_{i}\rho_{i}trig(X)$ or $c_{i}x+d_{i}\rho_{i}0$ (1)
with$i\in I,$ $\rho_{i}\in\{=, <, >\}$. Let$s$be $2\pi$plusthe maximal absolute value ofthe domainbounds
ofall lines $c_{i}x+d_{i}$ with $c_{i}\neq 0$ occurring in $\psi’$ and ofthe first components of all intersection
points between all lines $y=c_{i}x+d_{i}$ and $y=c_{j}’x+d_{j}’$ with $ci\neq 0$ orwith $c_{j}\neq 0$. Consider
onlythe region $[-s, s]$. Forevery atomic formula$\alpha_{i}$ in$\psi’$, the set $M_{\alpha_{i}}$ of reals defined by it in
$[-s, s]$ consists ofone interval
or
finitely manydisjoint intervals described in lemma2. So theset$M_{\psi’}$ of realsin $[-s, s]$ definedby$\psi’$ isafinite unionofintervals,whoseendpointsareamong
solutions ofthe equations
$c_{i}x+d_{i}=trig(x)$ or $c_{i}x+d_{i}=0$ (2)
obtained from the atomicformulas $\alpha_{i}$ in (1).
Lemma3
If
$\exists x(\psi’)$ holds in$\mathbb{R}$, thenitholds in $[-s, s]$.
ProofThisis animmediateconsequencefrom the definition of$s$ and lemma2. 1
Let$p_{1}<p_{2}<\ldots<p_{n}$ be the solutions of the equations (2) ordered in increasingorder. Then
the truth value of each atomic formula $\alpha_{j}$ of
$\psi’$ and hence of $\psi’$ remains invariant in each of
the intervals $(-\infty,p_{1}),$ $(p_{1},p_{2}),$ $..\mathrm{x}$
,
$(p_{n-1},p_{n}),$ $(p_{n}.\infty)$. So following the ideas in [10] in thepresent context, it suffices to determine the truth value of each $\alpha_{i}$ at the test$\mathrm{P}^{\mathrm{o}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{S}s}-,pi$, and
$p_{i}+\epsilon$, where $\epsilon$ is a positive infinitesimal. From the result of these evaluations one can then
determine the truth-value of$\psi’$ at all these points, andthus determine the structure ofthe setof
real values $x$ for which $\psi’(X)$ holds. In particularone can then decide, whetherornot $\varphi$ holds.
Moreprecisely,
one
obtains the exact numberofconnectedcomponents of the setofreal values$x$forwhich$\psi’(X)$ holds.
Twoproblems arise inthis procedure:
1. First,
one
needsto determine the order relations between the solutions ofthe equations (2).2. Second,
one
mustbe able toevaluate all atomic formulas$\alpha_{j}$ of$\psi’$ atthese solutions,closelytothe right ofthese solutions, and $\mathrm{a}\mathrm{t}-s$.
In contrast to the purely linear situation in [10], the solutions of the equations $c_{i}x+d_{i}$ $=$
$trig(x)(i\in I)$ are no longer given in a standard mathematical notation. So the evaluation of
$\alpha_{j}$ at or closely to theright ofthese solutions has to be done in a roundabout way. It turns out
that both problems can be
overcome
bycase
distinctions following lemma 2 and evaluation ofThe nexttwolemmasspecifythedetailsdepending
on
thecases
specified inlemma2.More-over, herewealsorestrict tothe
case
oftrig$(x)=\cos(x)$.Lemma 4 Let $\ell$ and $\ell’$ be a distinct straight lines
defined
by $y=cx+d$ and $y=c’x+d’$,respectively, where$C,$$C^{J}d,$$d’\in F$. Let$D$ be the interval $[-t, t]$ where
$t:= \max$
{
$- \frac{\pi}{2}+(n+1)\pi|n\in \mathrm{N}$ and $- \frac{\pi}{2}+n\pi<s$}.
$Let \mathcal{I}_{2n}=(-\frac{\pi}{2}+2n\pi, -\frac{\pi}{2}+(2n+1)\pi)$
for
integers$n$. Wedenoteasbefore
by$\xi_{2n,1}<\xi_{2n,2}$ thepotential solutions
of
theequation$\cos(x)=cx+d$ in$\mathcal{I}_{2n}\cap D$. Similarly wedenote by$\xi_{2n,1}’<$$\xi_{2n,2}’$ the potential solutions
of
the equation $\cos(x)=c’x+d’$ in$\mathcal{I}_{2n}\cap D$. We let $(\alpha, \beta)\in F^{2}$denote thepotential intersection point
of
the lines $\ell$ and$\ell’$. Notice that$\xi_{2n,1},$ $\xi_{2n,2},$ $\xi’2n,1’\xi_{2n}^{;},2$
are all
different from
$\alpha$, since $\cos(x)$ is strongly transcendental.If
the line $cx+d>0$ in aninterval, then itiscalledpositive in the interval.
If
oneof
the two lines $\ell$ and $\ell’$ has no intersection point with$\cos(x)$, the ordering
of
thesolutions$\xi_{i,j}$ are obvious
from
lemma 2. Hencewe only consider thecases where both two lineshave atleast one solution with $\cos(x)$ in $\mathcal{I}_{2n}$. Then the order relation among the solutions $\xi_{i,j}$
aredecidedas
follows:
1.
If
the two lines $\ell$ and $\ell’$ are bothpositive and have intersection points with$\cos(x)$ in $\mathcal{I}_{2n}$,
then let $\xi_{2n,1},$ $\xi_{2n},2$ be the intersection points with $\cos(x)$
for
the line $\ell$ and$\xi_{2n,1}’,$$\xi’2n,2$ be
those
for
the line $\ell’$. (Note that the case where $\cos(\alpha)=\beta$ is impossible, since$\cos(x)$ is
stronglytranscendental.)
$(a)$
Ifthere
is no intersectionpoint$of\ell$and$\ell’$ with$\alpha$in$\mathcal{I}_{2n}$ ortheintersection point$(\alpha, \beta)$has
$\alpha$ in$\mathcal{I}_{2n}$ and$\cos(\alpha)<\beta$, then wehave $\xi_{2n,1}’<\xi_{2n,1}<\xi_{2n,2}<\xi_{2n,2}’$
for
$c(2n\pi)+d>$$c’(2n\pi)+d’$ and$\xi_{2n,1}<\xi_{2n,1}’<\xi_{2n,2}’<\xi_{2n,2}$
for
$c(2n\pi)+d<c’(2n\pi)+d’$. $(b)$If
the intersectionpointof
$\ell$ and$l’$ has$\alpha$ in $\mathcal{I}_{2n}$ and$\cos(\alpha)>\beta$, then wehave $\xi_{2n,1}’<$
$\xi_{2n,1}<\xi_{2n,2}’<\xi_{2n,2}$
for
$c<c’$, and$\xi_{2n,1}<\xi_{2n,1}’<\xi_{2n,2}<\xi_{2n,2}’$for
$c>c’$.2.
If
the line$\ell$ is positive and has intersection points with$\cos(x)$ in $\mathcal{I}_{2n}$ and$\ell’$ has an
intersec-tion point$(\gamma, 0)$ withthe$x$-axis in $\mathcal{I}_{2n}$, let$\xi_{2n,1},$ $\xi_{2}n,2$ be the intersection pointswith $\cos(x)$
for
the line$\ell$and$\xi_{2n,1}’$ be that
for
the line1’.
(The casewhere the line$\ell’$ is positive and hasintersection pointswith $\cos(x)$ in $\mathcal{I}_{2n}$ and$\ell$ has an intersection point $(\gamma, 0)$ with the x-axis
in$\mathcal{I}_{2n}$ is similartothiscase.)
$(a)$ Consider the case $c’>0.\cdot$
If
$\cos(\alpha)<\beta$, then we $ha.ve\xi_{2n,1}<\xi_{2n,2}<\xi_{2n,1}’$.If
$\cos(\alpha)>\beta$, then wehave$\xi_{2n,1}<\xi_{2n,1}’<\xi_{2n,2}$.
$(b)$ Consider the case $c’<0$.
If
$\cos(\alpha)<\beta$, then we have $\xi_{2n,1}’<\xi_{2n,1}<\xi_{2n,2}$.If
$\cos(\alpha)>\beta$, then wehave$\xi_{2n,1}<\xi_{2n,1}’<\xi_{2n,2}$.
3.
If
the two lines $\ell$ and $\ell’$ have the intersection points$(\gamma, 0),(\gamma’, 0)$ with the $x$-axis in $\mathcal{I}_{2n}$,
respectively, let$\xi_{2n,1}$ be theintersection point with$\cos(x)$
for
theline$\ell$and$\xi_{2n,1}’$ be
thatfor
the line$\ell’$.
$(a)$
If
there isno intersection pointof
$\ell$ and$\ell’$ with$\alpha$ in $\mathcal{I}_{2n}$ orthe intersection point $(\alpha, \beta)$
$i$.
If
$c,$$c’>0$ then $\xi_{2n,1}<\xi_{2n,1}’$for
$c(2n\pi)+d>c’(2n\pi)+d’$ and$\xi_{2n,1}’<\xi_{2n,1}$for
$c(2n\pi)+d<c’(2n\pi)+d’$.
$ii$.
If
$c,$$c’<0$ then $\xi_{2n,1}’<\xi_{2n,1}$for
$c(2n\pi)+d>c’(2n\pi)+d’$ and $\xi_{2n,1}<\xi_{2n,1}’$for
$c(2n\pi)+d<c’(2n\pi)+d’$.$(b)$
If
the intersection point $\ell$and$\ell’$ has$\alpha$ in $\mathcal{I}_{2n}$ and $\cos(\alpha)>\beta$, then thefollowing cases
$occu\gamma.\cdot$
$i$.
If
$c,$$c’>0$ then $\xi_{2n,1}’<\xi_{2n,1}$
for
$c(- \frac{\pi}{2}+2n\pi)+d>c’(-\frac{\pi}{2}+2n\pi)+d’$ and$\xi_{2n,1}<\xi_{2n,1}’$
for
$c(- \frac{\pi}{2}+2n\pi)+d<c’(-\frac{\pi}{2}+2n\pi)+d’$.$ii$.
If
$c,$$c’<0$ then $\xi_{2n,1}<\xi_{2n,1}’$for
$c(- \frac{\pi}{2}+2(n+1)\pi)+d>c’(-\frac{\pi}{2}+2(n+1)\pi)+d’$and$\xi_{2n,1}’<\xi_{2n,1}$
for
$c(- \frac{\pi}{2}+2(n+1)\pi)+d<c’(-\frac{\pi}{2}+2(n+1)\pi)+d’$. $iii$.If
$cc’<0$ then $\xi_{2n,1}<\xi_{2n,1}’$for
$c<0,$$c’>0$ and$\xi_{2n,1}’<\xi_{2n,1}$for
$c>0,$$c’<0$.The interval$\mathcal{I}_{2n+1}$
for
aninteger$n$ istreatedin analogousway tothe interval$\mathcal{I}_{2n}$.Lemma5 Let $\xi_{2n,1}<\xi_{2n,2}$ be the potential solutions
of
the equation $cx+d=\cos(x)$ in$\mathcal{I}_{2n}\cap D$. Let the potential intersection point
of
two lines$\ell$ and1’
be $(\alpha, \beta)$. We only considerthe caseswhere both twolines haveatleastonesolution with$\cos(x)$ in$\mathcal{I}_{2n}\cap D$. Then the order
relation between$c’\xi 2n,i+d’$ and$\cos(\xi 2n,i)\mathcal{I}_{2}n$ isdecidedas
follows:
1.
If
the two lines$\ell$and1’ arebothpositive on$\mathcal{I}_{2n}$ and have intersection pointswith $\cos(x)$ in$\mathcal{I}_{2n}$, then let$\xi_{2n,1},$ $\xi_{2}n,2$ bethe intersectionpoints with $\cos(x)$
for
the line$\ell$.$(a)$
If
there isno intersectionpointof
1 and1’ with $\alpha$ in $\mathcal{I}_{2n}$ orthe intersectionpoint $(\alpha, \beta)$ has$\alpha$ in$\mathcal{I}_{2n}$ and$\cos(\alpha)<\beta$, then wehave $c’\xi_{2n},1+d’<\cos(\xi 2n,1),$ $C’(\xi_{2n,1}+\epsilon)+d’<$$\cos(\xi_{2}n,1),$ $c’\xi_{2n},2+d’<\cos(\xi 2n,2),$ $c^{J}(\xi 2n,2+\epsilon)+dJ<\cos(\xi_{2}n,2)$,
for
$c(2n\pi)+d>$$c’(2n\pi)+d’$ and$c’\xi_{2n},1+d’>\cos(\xi_{2}n,1),$ $c’(\xi_{2}n,1+\epsilon)+d’>\cos(\xi 2n,1),$$c\xi\prime 2n,2+d’>$
$\cos(\xi_{22}n,),$ $c(;\xi 2n,2+\epsilon)+d’>\cos(\xi_{2n,2})$
for
$c(2n\pi)+d<c’(2n\pi)+d’$.$(b)$
If
the intersection pointof
1and$\mathit{1}’$ has$a$ in $\mathcal{I}_{2n}$ and$\cos(\alpha)>\beta$, thenwe have$c’\xi_{2n},1+$ $d’<\cos(\xi 2n,1),$ $c’(\xi 2n,1+\epsilon)+d’<\cos(\xi_{2n,1}),$$c’\xi_{2n,2}+dJ>\cos(\xi 2n,2),$ $c’(\xi 2n,2+\epsilon)+$
$d’>\cos(\xi_{2n,2})$,
for
$c<c’,$ $c’\xi_{2n},1+d’>\cos(\xi_{2n,1}),$ $c’(\xi 2n,1+\epsilon)+d’>\cos(\xi_{2n,1})$, $C’\xi_{2n},2+d’<\cos(\xi 2n,2),$ $c’(\xi_{2n},2+\epsilon)+d^{J}<\cos(\xi 2n,2)$for
$c>c’$.2.
If
the line$\ell$is positive and hasintersectionpoints with $\cos(x)$ in$\mathcal{I}_{2n}$ and$\ell’$ has anintersec-tion point $(\gamma, 0)$ with the $x$-axis in$\mathcal{I}_{2n}$, let $\xi_{2n,1},$ $\xi_{2}n,2$ be theintersection points with $\cos(x)$
for
the line $\ell$. (The case where the line $\ell’$ is positive on$\mathcal{I}_{2n}$ and$\ell$ has an intersectionpoint$(\gamma, 0)$ with the$x$-axis in$\mathcal{I}_{2n}$ is similarto thiscase.)
$(a)$ Consider the case $c’>0.\cdot$
If
$\cos(\alpha)<\beta$, then we have $c’\xi_{2}n,1+d’<\cos(\xi_{2}n,1)$,$c’(\xi 2n,1+\epsilon)+d’<\cos(\xi_{2n,1})C’\xi_{2n},2+d’<\cos(\xi_{2}n,2),$ $C’(\xi 2n,2+\epsilon)+d’<\cos(\xi 2n,2)$.
If
$\cos(\alpha)>\beta$, then we have $c’\xi_{2}n,1+d’<\cos(\xi 2n,1),$ $c’(\xi_{2}n,1+\epsilon)+d’<\cos(\xi_{2}n,1)$$c’\xi_{22}n,+d’>\cos(\xi 2n,2),$ $C’(\xi 2n,2+\epsilon)+d’>\cos(\xi 2n,2)$.
$(b)$ Consider the case $c’<0$:
If
$\cos(\alpha)<\beta$, then we have $c’\xi_{2}n,1+d’<\cos(\xi_{2}n,1)$,$c’(\xi_{2}n,1+\epsilon)+d\prime \mathrm{c}<\mathrm{o}\mathrm{s}(\xi_{2n,1})_{C\xi 2n,2};+d^{;}<\cos(\xi_{22}n,),$ $C’(\xi_{2}n,2+\epsilon)+dJ<\cos(\xi 2n,2)$.
If
$\cos(\alpha)>\beta$, then we have $c’\xi 2n,1+d’<\cos(\xi_{2}n,1),$ $c’(\xi 2n,1+\epsilon)+d’<\cos(\xi_{2n,1})$ $c’\xi_{2n},2+d’>\cos(\xi 2n,2),$ $c’(\xi_{22}n,+\epsilon)+d’>\cos(\xi 2n,2)$.3.
If
thetwo lines and have the intersection points $(\gamma, 0),(\gamma’, 0)$ with$x$-axis in $\mathcal{I}_{2n}$respec-tively, let$\xi_{2n,1}$ be the intersection point with$\cos(x)$
for
the line$\ell$.$(a)$
If
there isno intersection pointof
$\ell$and$\ell’$ with$\alpha$ in $\mathcal{I}_{2n}$ or the intersection point $(\alpha, \beta)$ has $\alpha$ in$\mathcal{I}_{2n}$ and$\cos(\alpha)<\beta$, then the followingcases $ocCur.\cdot$
$i$.
If
$c,$$c’>0$then$c’\xi_{2n,1}+d’<\cos(\xi 2n,1),$ $c’(\xi_{2n,1}+\epsilon)+d’<\cos(\xi_{2n,1})$
for
$c(2n\pi)+$$d>c’(2n\pi)+d’$ and $c’\xi_{2}n,1+d’>\cos(\xi_{2n,1}),$ $c’(\xi 2n,1+\epsilon)+d’>\cos(\xi_{2}n,1)$
for
$c(2n\pi)+d<c’(2n\pi)+d’$.$ii$.
If
$c,$$c’<0$then$c’\xi_{2n},1+d’>\cos(\xi 2n,1),$ $c’(\xi_{2n,1}+\epsilon)+d’>\cos(\xi_{2n,1})$for
$c(2n\pi)+$ $d>c’(2n\pi)+d’$ and $c’\xi_{2}n,1+d’<\cos(\xi_{2}n,1),$ $c’(\xi 2n,1+\epsilon)+d’<\cos(\xi_{2n,1})$for
$c(2n\pi)+d<c’(2n\pi)+d’$.$(b)$
If
theintersection pointof
$\ell$and$\ell’$ has$\alpha$ in$\mathcal{I}_{2n}$ and$\cos(\alpha)>\beta$, then the followingcases
occur:
$i$.
If
$c,$$c’>0$ then $c’\xi_{2n,1}+d’>\cos(\xi_{2n},1),$ $C^{;}(\xi_{2n,1}+\epsilon)+d’>\cos(\xi_{2n,1})$
for
$c(- \frac{\pi}{2}+$$2n \pi)+d>c’(\frac{\pi}{2}+2n\pi)+d’$ and $c’\xi_{2}n,1+d’<\cos(\xi_{2n,1}),$ $c’(\xi_{2}n,1+\epsilon)+d’<$
$\cos(\xi_{2n,1})$
for
$c(- \frac{\pi}{2}+2n\pi)+d<c’(\frac{\pi}{2}+2n\pi)+d’$. $ii$.If
$c,$$c’<0$ then $c’\xi_{2}n,1+d’>\cos(\xi_{2}n,1),$ $C’(\xi_{2n,1}+\epsilon)+d’>\cos(\xi_{2}n,1)$
for
$c(- \frac{\pi}{2}+$$2(n+1) \pi)+d>c’(\frac{\pi}{2}+2(n+1)\pi)+d’$and$c’\xi_{2}n,1+d^{J}<\cos(\xi 2n,1),$ $C^{J}(\xi_{2}n,1+\epsilon)+d^{;}<$ $\cos(\xi_{2n,1})$
for
$c(- \frac{\pi}{2}+2(n+1)\pi)+d<c’(\frac{\pi}{2}+2(n+1)\pi)+d’$.$iii$.
If
$cc’<0$ then$c’\xi_{2n,1}+d’<\cos(\xi_{2}n,1),$ $c’(\xi_{2}n,1+\epsilon)+d’<\cos(\xi_{2}n,1)$.The interval$\mathcal{I}_{2n+1}$
for
integers $n$ is treatedin analogousway tothe interval$\mathcal{I}_{2n}$.Proof oflemma 4 and 5 These facts
are
an immediate consequence oflemma 2. Draw adiagram for eachcase. 1
Armed with the two lemmas we can now determine the exact number of non-empty
con-nected components of the set $M$ of all real values $x$, where $\psi’(X)$ holds. Since all coefficients
occurring in $\psi’$ are real algebraic numbers in $F$ and $\cos(x)$ is strongly transcendental we can
decide equations and order inequalities between the elements of $F$ and between the values of
$\cos(x)$for$x\in F$. Then all thecasedistinctionsmadeinthe twoproceedinglemmascanbe
eval-uated algorithmically. Sowe can determinealgorithmicallytheexactnumber of real solutions of
everyequation arising from
an
atomic subformulaof$\psi’$. Nextwe
can orderall these solutions bythe first lemma. Bythesecond lemmawe can evaluate the truth value ofeveryatomicsubformula
of$\psi’$ and hence of$\psi’$ at all thesepoints in $[-s, s]$, and closely to the right of all these points.
Fromtheseevaluations we obtainimmediately thenumber ofnon-empty connectedcomponents
of theset$M_{\psi’}$, andmoreoverthe type ofintervals thatthese componentsconstitute in increasing
order.
5
The mixed real-integer
case
In this section we consider a considerable extension of linear-exponential problems given
by
an
extension of the concept ofterm usedso
far. We admit asnew
operation symbol inour
extension language the integer-part operation [-] (compare [27]). Similar as in the language
addition, scalar multiplication by rational constants and the integer-part operation. Notice that
these operations maybe nested in an arbitrary way. Atomic formulas in the extended
sense
arethen ofthe form $t=0,$ $t>0,$ $t\geq 0$, for extended terms $t$. Formulas, closed formulas and
normal formulas in the extended
sense are
obtained from atomic formulasas
before with theadditional restriction that the quantifier
over
the mixed variable $x$ shouldrange now
onlyover
a bounded interval $[-S, S]$. By means of the integer-part operation we
are
now able to codealso quantifiersranging
over
the integers. Indeeda
quantifier $(\exists x\in \mathbb{Z})(\varphi)$can
be equivalentlyexpressedby the extended formula$\exists x$($x=[x]$ A
$\varphi$). So the problems that
can
be expressed bynormal formulasinthe extended
sense
arenow
mixed real-integer linear-trigonometric problems.We are now going to describe
a
decision procedurefor
normalformulas
in the extendedsense: By an argument similarto the one in section 2
we can
use the mixed real-integer linearquantifier eliminationof[27] toreduce the decision problem fornormal formulasto the
case
ofnormal formulaswithjust one existential quantifier
$\exists x(-s\leq x\leq s\wedge\psi(x))$ (3)
where $\psi(x)$ is quantifier-free. From the givenbound $S$
we can
compute an integer bound$T$ forthe absolutevalues of all extendedtermsand subterms$t(x)$ occurring in $\psi(x)\mathrm{f}\mathrm{o}\mathrm{r}-S\leq x\leq S$.
Next, we successively eliminate all
occurrences
ofthe integer-part operation from (3): Let $[t]$be aninnermost
occurrence
ofinteger-partoperation in $\psi(x)$. We write $\psi(x)$as
$\tau([t])$. Then(3)is equivalent to $_{i=-T}^{T}\exists x$($-s\leq x\leq S$ A $i\mathrm{l}\leq t(x)<(i+1)1$ A $\tau(i1)$) This decreases
the number of
occurrences
ofthe integerpart operation byone.
By iteration oftheprocedurealloccurrences
of theintegerpart operation canbe eliminatedfrom thegiven formula. Atthispointtheresulting formulacanbe decidedas in section4.
6
The
Tangent
case
So far the trigonometric function in normal formulas was essentially the sine
or
the cosinefunction. Whathappens ifwereplace this functionby the tangent function?
The obvious changes are the following: $tan(X)$ is defined only for $x \neq\frac{\pi}{2}+n\pi$ forinteger
$n,$ $tan(X)$ has period $\pi$, and $tan(X)$ is unbounded. We extend$tan(X)$ to
a
total function onIR
by setting $tan( \frac{\pi}{2}+n\pi)=0$. Sothe new tasksare thefollowing:
1. Compute the real zeros of the functions $f_{\mathrm{c},d}(x):=tan(X)+cx+d$where $c,$$d\in F$ and
determine the sign of$f_{c,d}(x)$ infinitesimallyto theright ofthese
zeros.
2. Eliminate the
occurrence
of the integer part operation from an univariate extended normalformulas.
For the first task we note to begin with that in each basic interval $\mathcal{I}_{n}$ the function $f_{c,d}(x)$
has exactly one zero, with the possible exception ofthe interval $\mathcal{I}_{m}$ containing the $\mathrm{z}\mathrm{e}\mathrm{r}\mathrm{o}-d/c$
of the line $cx+d$ in case $c\neq 0$. In the exceptional interval $\mathcal{I}_{m}$ the function has exactly one
zero if$c>-1$, or $c=-1$ and $d=0$, or$c<-1$ and the values of$tan( \pm acos(\frac{1}{\sqrt{-c}}+m\pi))$
have the same sign, and exactly three zeros otherwise. This follows from the following facts:
$f_{c,d(X)}’=(\cos(x\mathrm{I})^{-2}+c>0$ for $c>-1$. For $c<-1,$ $f_{c,d}’(X)$ has exactly two real zeros $\pm acos(\frac{1}{\sqrt{-c}}+n\pi)$ in each ofthe intervals$\mathcal{I}_{n}$. By the strong transcendence of$tan(X),$ $f_{c,d}(x)$
has
a
multiplezero
iff$c=-1$ and $d=0$. Using these factswe
can
solve the first task similarlyas
in section 4. Asa consequence we can
decidenormalformulas with trig$(x)=tan(X)$.We do not know how to solve the second task, since $tan(X)$ is unbounded. Instead
we
modify thetangent function by truncating its values at
some
fixedpositive integer $k$as
follows:Put $tan_{k}(x)=tan(X)$; if $|tan(X)|\leq k,$ $tan_{k}(x)=-k$ for$tan(X)<-k,$ $tan_{k(X})=k$ for
$tan(X)>k$. Thenthe second taskcanbe solved similarly
as
insection 6. So foreach fixed $k$weobtain
a
decisionprocedure for extendednormalformulas withtrig$(x)=tan_{k}(x)$.Again these decision procedures
can
be modifiedas
indicated before for quantifiers rangingover
a subfield of the field ofreal algebraic numbers.7
Deciding linear-trigonometric problems
over
subfields
So far we have consideredthe decision problemfor (extended) normal formulas, where all
quantifiers range
over
the field of real numbers. It is well-known that the truth value of linearformulas does not change when quantifiers are restricted to an arbitrary subfield of the reals
[10]. By [27] the same applies to mixed real-integer linear fornulas. For linear-trigonometric
problems thereis, however, abig differencebetweenvalidity in$\mathbb{R}$andin somesubfield$F$of the
real algebraicnumbers: Consider forexample the normal formula $\exists x(\cos(X)=x)$. Inthe reals
this formula is true, since the equation $\cos(x)=x$ has a unique real solution $\xi$ in the interval
$(0, \pi/2)$. By the strong trancendency of $\cos(x)$ this solution is transcendental, and hence the
formulaisfalsein $F$.
So
over
$F$theelimination oflinear variables from(extended)normal formulas remainsvalid,while thedecision ofunivariate normalformulas involvinga trigonometricfunction in section 4
maybecome incorrect.
In the following
we
sketch a modified decision algorithm that will be valid for arbitrarysubfields $F$ ofthe field of real algebraic numbers. As remarked aboveit suffices to modify the
decisionofunivariatenormal formulas $\exists x(\psi(x)$. Sincethe decision works bysubstituting finitely
many test points, it suffices to delete all test point that
are
not in $F$. These are exactly the testpoints of the form$\xi_{n,i}$ arising
as zeros
offunctions $f(x)=trig(x)+cx+d$ differentfromzero.
Notice, however,thatwe keepall the testpointof theform$\xi_{n,i}+\epsilon$, since they couldbe replaced
byanelement of$F$.
Thusinourexample $\exists x(\cos(X)=x)$ thedecisioninthereal lineuses astest$\mathrm{p}\mathrm{o}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{S}-1-2\pi$,
the solution $\xi$ ofthe equation $\cos(x)=x$, and the point $\xi+\epsilon$ for a positive infinitesimal $\epsilon$,
whereas in $F$ the testpoint $\xi$ is dropped. Hence the result of the decision is “true” in the reals
and“false” in$F$.
Remark
6
(Complexity of the decisionproblem) The complexityof
the decision problemfor
(extended)normal
formulas
hasnotbeen determinedyetexactly. We expectourdecision methodtorun in doubly exponential time
for
non-extended normalformulas
andin exponential timefor
8
Conclusions
We have shown that the decision problem forseveral classes offormulas involving real and
integervariables, linear polynomials and a trigonometric function is decidable. This is in
con-trast to theundecidability of the full elementarytheory of the reals with
a
specifictrigonometricfunction. The formulas may involve
one
mixed linear-trigonometric variable and severallin-ear
variables. The decision methoduses
linear quantifier elimination and a symbolic test pointmethod. It is quite explicit and implementable. Variants of it work
over
subfields of the realalgebraic numbers. In its approach it is similar to the decision method for linear-exponential
problems announced in [26]. We expect that theapproach
can
be generalizedto formulasinvolv-ing larger classes ofelementary functions.
Our decision method
can
be construedas
a modified quantifier elimination method, whenthe coefficients of terms
are
taken as parameters. Notice, however, that in thiscase
the outputformula willno longer benormal,but will include
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